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(D) By analyzing the circuit diagram,we can label the nodes. Let the potential at point $A$ be $V_A$ and at point $B$ be $V_B$.By tracing the connecti

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$64$

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(D) By analyzing the circuit diagram,we can label the nodes. Let the potential at point $A$ be $V_A$ and at point $B$ be $V_B$.By tracing the connections,we observe that all four capacitors are connected in parallel between points $A$ and $B$.For capacitors connected in parallel,the equivalent capacitance is given by $C_{eq} = C_1 + C_2 + C_3 + C_4$.Since all capacitors have the same capacitance $C = 16 \mu F$,we have:$C_{eq} = 16 \mu F + 16 \mu F + 16 \mu F + 16 \mu F = 64 \mu F$.Therefore,the equivalent capacitance between points $A$ and $B$ is $64 \mu F$.

Four capacitors,each of capacitance $16 \mu F$,are connected as shown in the figure. The equivalent capacitance between points $A$ and $B$ is: . . . . . . (in $\mu F$).

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